Question

mercury circles the sun once every 0.241 years. what is its semi - major axis length as it orbits the sun? (1 point)
0.39 au
0.12 au
0.241 au
0.058 au

Explanation:

Step1: Apply Kepler's third law

Kepler's third law is \(T^{2}=a^{3}\) (where \(T\) is the orbital period in years and \(a\) is the semi - major axis in astronomical units (AU)). Given \(T = 0.241\) years.

Step2: Solve for \(a\)

We have \(a=\sqrt[3]{T^{2}}\). Substitute \(T = 0.241\) into the formula: \(a=\sqrt[3]{(0.241)^{2}}\).
First, calculate \((0.241)^{2}=0.058081\). Then, find the cube root of \(0.058081\). \(\sqrt[3]{0.058081}\approx0.39\)

Answer:

0.39 AU